Name______

Date ______Pd______

NOTES: LIMITING AND EXCESS REACTANTS (There’s Only 3 Problems at the end of these notes)

Limiting Reactantis the reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed (used up).

Excess Reactantis the reactant in a chemical reaction that remains when a reaction stops when the limiting reactant is completely consumed. The excess reactant remains because there is nothing with which it can react. Think of the excess reactant is left-overs (just because someone cooked three pies, does not mean that all three pies are eaten).

How to find the limiting reactant:The following scenario illustrates the significance of limiting reactants. In order to assemble a car, 4 tires and 2 headlights are needed (among other things). In this example, imagine that the tires and headlights are reactants while the car is the product formed from the reaction of 4 tires and 2 headlights. If you have 20 tires and 14 headlights, how many cars can be made?

Given(s) and unknown→ / 20 / 14 / X
Balanced equation → / 4 Tires / + 2 Headlights / = / 1 Car

We now have to determine which reactant (tires or headlights) we will run out of first. We actually have two givens and one unknown, so we have two problems to work. After identifying one of the givens and the unknown, write a conversion to connect them:“4 tires=1 car”and perform your dimensional analysis.

20 tires / 1 car / = 5 cars
4 tires

Now you have to solve with the second given; write a conversion to connect them: “2 headlights: 1 car” and perform the second dimensional analysis.

14 headlights / 1 car / = 7 cars
2 headlights

Since we can only make five product cars with the amount of reactants provided, the tires will LIMIT the reaction. The other reactant (the headlights) is more than we need for the five cars, so we have left-over headlights making the headlights the EXCESS reactant.In this case, the excessive amount is determined by simple arithmetic: for five cars we need 10 headlights, we have 14 – 10 = 4 headlights left-over.

The same logic determines limiting and excess reactants (sometimes the word “reagent” is used for the word “reactant” – don’t let that throw you) in real world chemistry. Consider the following problem and solution:

A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen producing nitrogen monoxide and water. Which is the limiting reactant and how much excess reactant remains after the reaction has stopped?

Step 1: Write a balanced equation 4 NH3(g) + 5 O2(g)4 NO(g) + 6 H2O(g)

Step 2: Identify the given(s) and unknown. It is clear that the unknown is not specifically given, but does it really matter which product you pick to work? Not really, because if the reaction stops, it does not matter which one you pick to use as your unknown – BUT BE SURE TO USE THE SAME ONE in both of your analyses.

Given and unknown → / 2.00 g / 4.00 g / X g
Balanced equation → / 4NH3 / + 5O2 / → 4NO / + 6 H2O

Step 3A: Write the conversions for the dimensional analysis using the first given (2.00 g NH3).

4 mol NH3 = 4 mol NO

1 mol NH3 = [14.007 + 3(1.008)] = 17.031 g NH3

1 mol NO = [14.007 + 15.999] = 30.006 g NO

Step 4A: Dimensional Analysis

2.00 g NH3 / 1 mol NH3 / 4 mol NO / 30.006 g NO / = 3.52 g NO
17.031 g NH3 / 4 mol NH3 / 1 mol NO

Step 3B: Write the conversions for the dimensional analysis using the second given (4.00 g O2).

5 mol O2 = 4 mol NO

1 mol O2 = [2(15.999)] = 31.998 g O2

1 mol NO = [14.007 + 15.999] = 30.006 g NO

Step 4B: Dimensional Analysis

4.00 g O2 / 1 mol O2 / 4 mol NO / 30.006 g NO / = 3.00 g NO
31.998 g O2 / 5molO2 / 1 mol NO

Since the result of the second problem is smaller, O2 is the limiting reactant.

To determine the amount of EXCESS NH3 we need to find out how much was actually used in the reaction with 4.00 g O2. So . . . .

Given and unknown → / X g / 4.00 g
Balanced equation → / 4NH3 / + 5O2 / → 4NO / + 6 H2O

Conversions:

4 mol NH3 = 5 mol O2

1 mol NH3 = 17.031 g NH3

1 mol O2 = 31.998 g O2

Dimensional analysis:

4.00 g O2 / 1 mol O2 / 4 mol NH3 / 17.031 g NH3 / = 1.70 g NH3
31.998 g O2 / 5 mol O2 / 1 mol NH3

We started with 2.00 g NH3 but were only able to use 1.70 g NH3, so 0.30 g NH3 is in excess.

Now it is your turn to solve some Limiting/Excess Problems. Use your own paper to solve these problems, you MUST show your dimensional analysis.

1. Calculate the mass of magnesium oxide possible if 2.40 grams of magnesium react with 10.0 grams of oxygen. The balanced equation is 2Mg + O2 →2MgO

2. What is the limiting reagent if 76.4 grams of C2H3Br3 were reacted with 49.1 grams of O2 producing carbon dioxide, water and bromine? __ C2H3Br3 + ___O2 → ___CO2 + ___H2O +___ Br2

3. What is the limiting reactant if 28.7 grams of silicon dioxide react with 22.6 grams of H2F2?

The equation is: ___SiO2 + ___H2F2 →___SiF4 + ___H2O.

Hint: To find the limiting reactant, you must do 2 dimensional analysis problems each with the 2 different amounts given in the problem. Then, you must convert both of them to the same chemical and unit so that you can see which one makes the least amount. The original value that makes the least amount IS your limiting reactant. (keyword : REACTANT). For example, on number two you should do two problems and convert both to grams of H2O, or grams of Br2, or grams of CO2.It really does not matter as long as you convert BOTH to the same one so that you can compare them.